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《MATLAB/Simulink电力系统建模与仿真》中94页中的2机5节点电力系统潮流计算仿真,我参数完全按照书中设置,但调试有问题,Update Load Flow之后,得出发电机的三相电压不对称,数据如注释所示。哪位高手知道为啥?
9 [; s y- g2 y& m$ q( Y注:
: o$ O/ x4 r3 m$ a, c9 D6 }3 xMachine: G2
- Z+ o8 N* L& ]# _3 ONominal: 100 MVA 10.5 kV rms 4 d: A* A2 E1 _
Bus Type: Swing bus
, P* ~( h) b1 q3 R! _' \. j! K) |Uan phase: -40.27°
: Z4 J9 G& {9 X" F' a; oUab: 52029 Vrms [4.955 pu] -10.27° $ V) h2 r: E, g3 }- P* t! t ]
Ubc: 48895 Vrms [4.657 pu] 157.70° & h7 N& W6 b9 ^# K( h
Uca: 11025 Vrms [1.05 pu] -122.71°
( D9 n) |6 p X H, v2 L, L! zIa: 16777 Arms [3.051 pu] -9.91° & J ]( B7 H, L4 d& W7 f7 t
Ib: 16777 Arms [3.051 pu] -129.91° * y7 G* d4 b; t; P! F0 z1 w1 g
Ic: 16777 Arms [3.051 pu] 110.09° 5 [) D, K9 e$ d
P: 1.3046e+009 W [13.05 pu] 3 g: e2 B _% Q5 f6 ^
Q: -7.6415e+008 Vars [-7.642 pu] : S. i8 W' z0 a- t
Pmec: 1.3079e+009 W [13.08 pu] : x1 K( l# W0 r& `. o: C3 _% b+ Q
Torque: 8.3264e+006 N.m [13.08 pu] # C) e. x# R( R5 S) `; F, c
Vf: 6.067 pu
$ F' W7 \9 u- L q, E! H8 T7 N
9 U9 G/ Y, S# H, ^- fMachine: G1 : R6 g3 K5 I3 a6 c0 f! |
Nominal: 100 MVA 10.5 kV rms
0 a$ r! F+ b. n3 ~& M, R- s4 wBus Type: P & V generator
- ?1 o3 `: L& n6 q' o* j t) _9 IUan phase: -40.45°
+ |, Z! x1 [/ G; t3 z) @Uab: 55176 Vrms [5.255 pu] -10.45°
8 g. d) G0 I; }) l( N; S4 aUbc: 50638 Vrms [4.823 pu] 158.65°
7 {. P* D1 Q# C9 s! f. b! H' O N9 [Uca: 11025 Vrms [1.05 pu] -130.09° S5 x' u5 j& U3 Z: B7 a& v
Ia: 17338 Arms [3.153 pu] -21.32° ! v2 K. J" c% j$ ~, ^ h+ P. O
Ib: 17338 Arms [3.153 pu] -141.32°
: T. P" l& l4 ]Ic: 17338 Arms [3.153 pu] 98.68°
" \; D4 }5 }5 N+ S# C3 IP: 1.5654e+009 W [15.65 pu] M( Z0 u$ J$ ^5 K
Q: -5.4297e+008 Vars [-5.43 pu]
- x, R6 b' c% H) g6 b8 Y$ sPmec: 1.569e+009 W [15.69 pu] 5 w1 J3 k( `2 E) W' h
Torque: 9.9887e+006 N.m [15.69 pu] 6 n/ p# V$ K7 ?& Q K. G( c/ _
Vf: 6.7898 pu % W8 v" L, Z5 i$ L4 O0 K H! @
" J/ }" @6 p: p) K q
Machine: Load3 * A, {; M& F0 \4 ~
Nominal: 115 kV rms 9 h! \! q9 b( V& d4 c* l+ B: z
Bus Type: P & Q load
& u- ?2 w: n+ o( BUan phase: -46.14° * S4 U; c) u$ _: [0 z/ c" z
Uab: 6.2602e+005 Vrms [5.444 pu] -16.14°
* M, b. @; f8 M" zUbc: 6.1132e+005 Vrms [5.316 pu] 161.43°
% p/ A0 x1 }( mUca: 30061 Vrms [0.2614 pu] -136.62° ) H/ _5 p( n3 ]. G2 M+ r+ p% f; m
Ia: 300.16 Arms -102.68°
$ g- M7 r/ K: B$ \% |9 E, j8 nIb: 300.16 Arms 137.32° ' m( }0 S, N* n( O( r; t5 ~/ d
Ic: 300.16 Arms 17.32°
2 U& o* I& c( t' S3 L& J& FP: 1.7944e+008 W ; Q/ J7 A. e# V2 h2 \
Q: 2.7153e+008 Vars ) p$ A. z- i1 |- R
. M3 Y# b) {1 p3 C* k% NMachine: Load2 3 W/ R4 w( X" v" }8 H! {/ N
Nominal: 115 kV rms O: u" ?" D/ x, P, n' ^
Bus Type: P & Q load : x" _$ N+ U; h4 j8 X* @
Uan phase: -41.18° - X$ K) J0 J# S, Z- X* v$ a/ e1 n$ e
Uab: 6.0512e+005 Vrms [5.262 pu] -11.18°
7 q/ E% N6 i u( q0 }Ubc: 5.7431e+005 Vrms [4.994 pu] 157.92°
1 Z" c9 Z- ^+ a/ v. O {4 S) yUca: 1.1609e+005 Vrms [1.009 pu] -121.95°% h7 S7 o4 ~2 s0 Q
Ia: 798.9 Arms -92.72° ; S& p3 |# J+ H; U
Ib: 798.9 Arms 147.28°
1 j( z4 c6 b* Y Y. pIc: 798.9 Arms 27.28° " U9 t$ V4 X, e: K: m1 E6 [2 e# z
P: 5.2087e+008 W + Q) J$ ]7 k* O6 ^; m; T
Q: 6.5561e+008 Vars
* n% }% N: u" | F, p, t, _
9 i# c! X0 Z* H4 p4 L# r) TMachine: Load1
[7 R+ x% L. u1 E; ~6 eNominal: 115 kV rms 9 |! J% K5 o& g" w
Bus Type: P & Q load ( k2 b: `! C3 r% C2 S
Uan phase: -40.94° ; D) `0 j7 t1 [6 k- q1 i
Uab: 6.3759e+005 Vrms [5.544 pu] -10.94° : F( ~/ y: W8 ?) Z3 X3 u
Ubc: 5.8926e+005 Vrms [5.124 pu] 158.63°
+ W8 L% n3 J o' ]+ Y- ^% sUca: 1.2147e+005 Vrms [1.056 pu] -129.49°) S7 p7 g9 ~* f6 f; B' o" H$ \# P
Ia: 433.21 Arms -97.84°
* V( m/ f2 q! f" Z- r; oIb: 433.21 Arms 142.16° 7 J% J$ j, [1 W3 e: x' d
Ic: 433.21 Arms 22.16°
4 _7 S) S5 O* \* N/ RP: 2.6123e+008 W
& E% f) w0 A% A7 X8 l" `Q: 4.0079e+008 Vars |